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3 years ago

Help me please ┐(‘～`;)┌

Chemistry

2 answers:

slava [35]3 years ago

8 0

Answer: (2)react completely in the reaction

limiting reactant determines the number of moles in a chemical reaction

excess reactant is the reactant with an excess number of moles

Helga [31]3 years ago

4 0

Answer:

<h3>Limiting reagent in chemical reaction is that reactant which Left some amount unreacted after the completion of reaction.</h3>

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How many grams of h3po3 would be produced from the complete reaction of 93.2 g p2o3

ki77a [65]

Answer: 137.76 g

Explanation:

3 0

2 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: The mass of glycine that can be dissolved is $1.3\times 10^2g$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

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3 years ago

Which of the following is not an example of a molecule A. Mn B. KOH C. O₃ D. H₂S

goldenfox [79]

look at the chemical tables. but i believe it is A

3 0

3 years ago

Read 2 more answers

Thomas calculates the volume of an object by multiplying the measurements 1.74 cm x 2.3 cm x 3.09 cm. Using significant figures,

Gekata [30.6K]

Answer:

12 cm³

Explanation:

Always round to the place of the number with the least amount of significant figures.

1.74 and 3.09 have three significant figures. 2.3 has two significant figures. Therefore, the answer will be rounded to two significant figures.

1.74 cm × 2.3 cm × 3.09 cm = 12.36618 -> 12 cm³

(The 3 in the tenths place is less than 5 so drop it and all the numbers to the right of it.)

Hope that helps.

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3 years ago

Cream of tartar has a pH that is (acidic or basic)

son4ous [18]

It is acidic because it has the Carboxylic Acid.

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3 years ago

Help me please ┐(‘～`;)┌​

Help me please ┐(‘～`;)┌